已知f(z)=2z+2-i,z0=1+2i,f(z0-z1)=6-3i,z∈c,求复数z1,f（z0+z1Ι）.

Question

已知f(z)=2z+2-i,z0=1+2i,f(z0-z1)=6-3i,z∈c,求复数z1,f（
z0+z1Ι）.

Answer 1

let
z1=a+bi
z0-z1= (1-a)+ (2-b)i
f(z)=2z+2-i
f(z0-z1) =6-3i
2(1-a)+ 2(2-b)i + 2-i =6-3i
(4-2a)+ (3-2b)i =6-3i
=>
4-2a =6 and 3-2b=-3
a=-1 and b=3
z1= -1 +3i
z0 +z1i
=1+2i + (-1+3i)i
=-2+i
f(z0 +z1i) =2(-2+i ) + 2-i
=-2+i